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This is raised by a recent question occurring in combinatorial geometry. It is about a sort of tractrix, but instead of a line, the pulling end moves along a circle of radius $r>\frac12$ (typically around $0.8$), starting with a chord if length 1. Imagine this chord horizontally on the bottom and take its center as the origine of a cartesian coordinate system. If the circle has center $(0,h)$, we have $h^2+\frac14=r^2$, so only one of $h$ and $r$ is needed to define the tractrix.
enter image description here

The tractrix is approximately the green line in the picture, where all blue segments (the tangents) have unit length.
Here is what I have got so far :

For a point $(x,y)$ on the tractrix where it has slope $f’$, we have 3 equations for the point $(u,v)$ where the tangent intersects the circle on the right:
$(i)\ \ \ \ v-y=(u-x)f’\ $ (equation of the tangent)
$(ii) \ \ \ u^2+(h-v)^2=r^2\ $ (intersection with the circle), equivalently $u^2+v^2-2hv-\frac14=0$. $(iii)\ \ (u-x)^2+(v-y)^2=1\ $ (constant length of the tangent between tractrix and circle).

Eliminating $u$ and $v$ yields the following differential equation for the tractrix: $$x+\sqrt{1-s^2}=\sqrt{2h(y+s)-(y+s)^2-\frac14}$$ where $$s :=\frac{f'}{\sqrt{1+f’^2}}.$$ (Thus $s$ is the sine of the slope angle.)

I don’t think there is a closed form of the tractrix equation. But is there a way to determine at which point $(x,y)$ the tangent is vertical? I'd expect it to be a not-too-involved function of $h$.

Note that, unless $r$ is too small for reaching the vertical position at all, the tractrix will carry on winding beyond the vertical for a finite time. That is, under the assumption that at each moment, the segment is tangent to the tractrix at its (the segment's) endpoint. (Imagine $r$ a bit smaller than in the picture, such that the vertical tangent of the tractrix coincides with the $y$-axis. From there on, the top of the blue segment cannot move further to the left without "breaking the smoothness" of the tractrix. I think that for each $r$, the movement will eventually arrive at such a point, which we will naturally consider as the endpoint of the tractrix.)

But where is that point? I have no idea how far it can go if $r$ is big, but I don't think it will go further than becoming horizontal again. All this is easier to perceive if we keep $r$ constant and require the 'rotating' segment of length $\epsilon$ instead of unit length. So:

Where does the tractrix stop if $\epsilon\to0$?

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    $\begingroup$ What's perhaps even tricker is to say "tricky tractrix" ten times fast. $\endgroup$ Nov 8, 2015 at 18:27
  • $\begingroup$ Perhaps it would be easier to rotate $\pi\over2$ and search for $f'=0$. $\endgroup$
    – Jeff Strom
    Nov 8, 2015 at 19:59
  • $\begingroup$ @JeffStrom I think that would not make a big difference, just swap $x$ and $y$. $\endgroup$
    – Wolfgang
    Nov 8, 2015 at 20:31

1 Answer 1

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In fact, using the moving frame, it is easy explicitly to solve the equations and get the formula for the slope $\tan\bigl(\theta(s)\bigr)$ as a function of arc-length along the curve. However, one sees immediately that it is a transcendental function, so explicitly solving for the value of $s$ for which $\theta(s) = \tfrac12\pi$ is not going to be easy.

Here are the steps: Let $$ X(s) = \pmatrix{x(s)\\y(s)},\qquad \text{with}\qquad X(0) = \pmatrix{-\tfrac12\\-\sqrt{r^2-\tfrac14}} $$ be the arclength parametrization of the curve, and set $$ E_1 = X'(s) = \pmatrix{\cos\bigl(\theta(s)\bigr)\\\sin\bigl(\theta(s)\bigr)} \quad\text{and}\quad E_2 = \pmatrix{-\sin\bigl(\theta(s)\bigr)\\\cos\bigl(\theta(s)\bigr)}. $$ We then have structure equations $X'(s) = E_1(s)$, $E_1'(s) = \kappa(s)E_2(s)$, and $E_2'(s) = -\kappa(s)E_1(s)$, where $\kappa(s) = \theta'(s)$. We also have $\theta(0) = 0$, by the above normalization.

We are requiring that $\bigl|X(s)+E_1(s)\bigr| = r$. (This is the tractrix equation.)

Let us set $$ X(s) = u(s)\ E_1(s) + v(s)\ E_2(s), $$ where the initial conditions imply $u(0) = -\tfrac12$ and $v(0) = -\sqrt{r^2-\tfrac14}$.

Then the above structure equations imply, since $X'(s) = E_1(s)$, that $$ u'(s) = 1 + \kappa(s) v(s)\qquad\text{and}\qquad v'(s) = -\kappa(s) u(s). $$

Moreover, the tractrix equation implies that $(u(s)+1)^2 + v(s)^2 = r^2$. Differentiating this equation and using the above differential equations then yields $$ 1 + u(s) + \kappa(s)v(s) = 0, $$ which implies $u'(s) = -u(s)$, which, with the initial condition above implies $$ u(s) = -\tfrac12\,e^{-s} $$ and then, consequently, that $$ v(s) = -\sqrt{r^2 - \bigl( 1-\tfrac12\,e^{-s} \bigr)^2}. $$ Moreover, we have $$ \theta'(s) = \kappa(s) = -\frac{(1+u(s))}{v(s)} = \frac{1-\tfrac12\,e^{-s}}{\sqrt{r^2 - \bigl( 1-\tfrac12\,e^{-s} \bigr)^2}}, $$ i.e., $$ \theta(s) = \int_0^s \frac{2-e^{-\sigma}}{\sqrt{4r^2-\bigl( 2-e^{-\sigma}\bigr)^2}}\,\mathrm{d}\sigma. $$ (Note, by the way, that this is an elementary integral, but the integral is a transcendental function.)

Armed with the knowledge of $u(s)$, $v(s)$, and $\theta(s)$, the above formulae give an explicit arc-length parametrization of the tractrix. Note, that, of course, we must have $r>\tfrac12$ or the equations (and the condition) don't make sense.

When $r\ge1$, these formulae exist for all $s\ge0$ and, of course, $u(s)$ goes to zero as $s\to\infty$, while $v(s)$ goes to $-\sqrt{r^2-1}$, i.e., the tractrix asymptotically approaches a circle of radius $\sqrt{r^2-1}$. Moreover, not surprisingly, $\theta(s)$ increases without bound as $s\to\infty$, so there is a first value of $s$ for which $\theta(s) = \tfrac12\pi$.

When $\tfrac12<r<1$, the curvature of the curve goes to infinity in finite arclength, in fact, when $s = -\log\bigl(2(1{-}r)\bigr)>0$. At this point, the curve will have a cusp, and the tangent line will point directly towards the origin. In fact, when $r < 0.724651057$, the (strictly increasing) function $\theta$ will not actually reach $\tfrac12\pi$ in the interval $0\le s\le -\log\bigl(2(1{-}r)\bigr)$, so the curve does not turn vertical in this range before its curvature goes to infinity.

Addendum: Continuing the curve past the cusps

It turns out that there is a reasonable way to continue the curve past the cusps, and this gives a more reasonable picture. The way to do this is to realize that one can reparametrize the curve smoothly (not by arclength) in a smooth way as follows: Consider the equations $$ \mathrm{d}u = -u\,\mathrm{d}s,\quad \mathrm{d}v = \frac{u(u+1)}{v}\,\mathrm{d}s,\quad \mathrm{d}\theta = -\frac{(u+1)}{v}\,\mathrm{d}s\,. $$ Setting $\mathrm{d}s = v\,\mathrm{d}t$, these can be rewritten as $$ \mathrm{d}u = -uv\,\mathrm{d}t,\quad \mathrm{d}v = u(u+1)\,\mathrm{d}t,\quad \mathrm{d}\theta = -(u+1)\,\mathrm{d}t\,. $$ Now, regarding $(u,v,\theta)$ as functions of $t$, we get the formula for $X(t)$ in the form $$ X(t) = u(t)\ \pmatrix{\cos\bigl(\theta(t)\bigr)\\\sin\bigl(\theta(t)\bigr)} + v(t)\ \pmatrix{-\sin\bigl(\theta(t)\bigr)\\\cos\bigl(\theta(t)\bigr)}. $$ Note that $(\dot u,\dot v) = \bigl(-uv,u(u{+}1)\bigr)$ defines a vector field on the circle $(u{+}1)^2+ v^2 = r^2$, and, when $\tfrac12<r<1$, this vector field has no zeros. Hence $u$ and $v$ are smooth periodic functions of $t$ of some period $T(r)$, and $\theta$ satisfies $\theta(t+T(r)) = \theta(t) + P(r)$, where $$ P(r) = \int_0^{T(r)} (1+u(t))\,\mathrm{d}t. $$ (The cusps occur where $v$ vanishes.) In fact, it is easy to compute these period integrals when $r^2<1$. One finds that $$ T(r) = \int_{1-r}^{1+r}\frac{2\,\mathrm{d}\rho}{\rho\sqrt{r^2-(1{-}\rho)^2}} = \frac{2\pi}{\sqrt{1-r^2}} $$ and $$ P(r) = \int_{1-r}^{1+r}\frac{2(1{-}\rho)\,\mathrm{d}\rho}{\rho\sqrt{r^2-(1{-}\rho)^2}} = \frac{2\pi\bigl(1-\sqrt{1-r^2}\bigr)}{\sqrt{1-r^2}}\,. $$ In particular, note that, when $\sqrt{1-r^2}$ is rational, the curve closes periodically, with a finite number of cusps. The length of the curve between consecutive cusps is $$ S = \log\left(\frac{1+r}{1-r}\right)\,. $$

A further remark on integration of the tractrix

In fact, by a change of variable, one can integrate the equations completely when $r^2<1$. The equation $(u{+}1)^2+v^2=r^2$ together with the differential equations $$ \mathrm{d}u = -uv\,\mathrm{d}t,\quad \mathrm{d}v = u(u+1)\,\mathrm{d}t,\quad \mathrm{d}\theta = -(u+1)\,\mathrm{d}t\,. $$ show that we can actually write $$ (u,v) = \bigl(r\,\cos 2\phi -1,\ r\,\sin 2\phi) $$ for some function $\phi$ on the curve. From the first two differential equations, this implies $\mathrm{d}t = -2\,\mathrm{d}\phi/(1-r\,\cos 2\phi)$, so $$ \mathrm{d}\theta = -(u+1)\,\mathrm{d}t = \frac{2r\,\cos 2\phi\,\mathrm{d}\phi}{(1-r\,\cos 2\phi)} $$ and this integrates to give $$ \theta(\phi) = 2\left( \frac{\arctan\bigl(\sqrt{\frac{1{+}r}{1{-}r}}\,\tan\phi\bigr)}{\sqrt{1-r^2}}-\phi\right). $$ Hence, $$ X(\phi) = \pmatrix{r\,\cos\bigl(2\phi{+}\theta(\phi)\bigr)-\cos(\theta(\phi))\\ r\,\sin\bigl(2\phi{+}\theta(\phi)\bigr)-\sin(\theta(\phi))}, $$ As a result, it follows that, when $\sqrt{1-r^2}$ is rational, the curve not only closes periodically, but it is algebraic. I suspect that this was known classically.

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  • $\begingroup$ Thank you very much! I guess that is about the best one can say. Now as you have been able to find numerically the radius which makes the tractrix stop at $\frac\pi2$, can you also find numerically for which $r$ we have $x+y=r$ for the cusp $(x,y)$? This would be the less wasteful way to keep space for inserting a whole quandrant in my original construction: the bottom end of the vertical tangent lying on the diagonal line of slope $-1$ through the center of the circle (as BTW is roughly the case in my drawing above). $\endgroup$
    – Wolfgang
    Nov 9, 2015 at 9:48
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    $\begingroup$ @Wolfgang: Yes, that value turns out to be about $r=0.8250033$. $\endgroup$ Nov 9, 2015 at 13:03
  • $\begingroup$ Thank you again! I have incorporated that in the other thread. $\endgroup$
    – Wolfgang
    Nov 9, 2015 at 13:41
  • $\begingroup$ @Wolfgang: The value of $r$ for which the curve first turns vertical exactly when it reaches the line $x+y = 0$ is $r\approx 0.8250033$. However, that curve does not have a cusp at that point; the cusp happens further along. By the way, I hope that you have noticed that I have normalized so that the 'handle' drawing the point along is moving along the curve $x^2+y^2 = r^2$, not $x^2 + (y-h)^2 = r^2$, as you have drawn it in your diagram. $\endgroup$ Nov 9, 2015 at 13:41
  • $\begingroup$ Yes I know about the cusp happening later. Do you think, like somebody had suggested in between in a comment now deleted, that the optimal solution of the original problem MUST use an 'entire' tractrix (i.e. ending at a cusp)? Re your 2nd remark: Yes I have seen that your $X(0)$ is shifted by $-h$ compared with my coordinates. But does that mean your results refer to a different construction? I hope not! $\endgroup$
    – Wolfgang
    Nov 9, 2015 at 14:06

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